Fourier transform of $L^2$ function

Question

Let us the definition of Fourier transform $$\hat f(\lambda) = \int_{-\infty}^\infty f(t) \exp(- i \lambda t) dt$$ How do I change this expression if $f(t)\in L^2[-\pi,\pi]$? $$\hat f(\lambda) = \int_{-\pi}^\pi f(t) \exp(- i \lambda t) dt\ \ ?$$

Answer 1

It depends on what you are after: what do you want to do with the transform? What you described is a reasonable thing to do; it amounts to extending $f$ by zero outside of the interval $[-\pi,\pi]$ and applying the usual formula on the line. A possible drawback: if $f$ is not zero at $\pm \pi $, the steep drop-off will result in low accuracy near the endpoints (Gibbs phenomenon).

Another approach is to expand $f$ into Fourier series. That would be $$\hat f(n) = \frac{1}{2\pi}\int_{-\pi}^\pi f(t) \exp(- i n t) dt,\qquad n\in\mathbb Z$$ Essentially the same formula, but you only compute countably many coefficients, and reconstruct $f$ with the formula $$f(t) = \sum_{n\in\mathbb Z} \hat f(n) \exp(i n t)$$ which is a convergent series in the $L^2$ sense (and also pointwise at almost every point).

The series works particularly well when $f(\pi)=f(-\pi)$. Otherwise the Gibbs phenomenon will show up again.

The placement of $2\pi$ is a matter of taste, by the way.